0.000 000 021 979 552 668 138 406 918 015 Converted to 64 Bit Double Precision IEEE 754 Binary Floating Point Representation Standard

Convert decimal 0.000 000 021 979 552 668 138 406 918 015(10) to 64 bit double precision IEEE 754 binary floating point representation standard (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

What are the steps to convert decimal number
0.000 000 021 979 552 668 138 406 918 015(10) to 64 bit double precision IEEE 754 binary floating point representation (1 bit for sign, 11 bits for exponent, 52 bits for mantissa)

1. First, convert to binary (in base 2) the integer part: 0.
Divide the number repeatedly by 2.

Keep track of each remainder.

We stop when we get a quotient that is equal to zero.


  • division = quotient + remainder;
  • 0 ÷ 2 = 0 + 0;

2. Construct the base 2 representation of the integer part of the number.

Take all the remainders starting from the bottom of the list constructed above.

0(10) =

0(2)


3. Convert to binary (base 2) the fractional part: 0.000 000 021 979 552 668 138 406 918 015.

Multiply it repeatedly by 2.

Keep track of each integer part of the results.

Stop when we get a fractional part that is equal to zero.


  • #) multiplying = integer + fractional part;
  • 1) 0.000 000 021 979 552 668 138 406 918 015 × 2 = 0 + 0.000 000 043 959 105 336 276 813 836 03;
  • 2) 0.000 000 043 959 105 336 276 813 836 03 × 2 = 0 + 0.000 000 087 918 210 672 553 627 672 06;
  • 3) 0.000 000 087 918 210 672 553 627 672 06 × 2 = 0 + 0.000 000 175 836 421 345 107 255 344 12;
  • 4) 0.000 000 175 836 421 345 107 255 344 12 × 2 = 0 + 0.000 000 351 672 842 690 214 510 688 24;
  • 5) 0.000 000 351 672 842 690 214 510 688 24 × 2 = 0 + 0.000 000 703 345 685 380 429 021 376 48;
  • 6) 0.000 000 703 345 685 380 429 021 376 48 × 2 = 0 + 0.000 001 406 691 370 760 858 042 752 96;
  • 7) 0.000 001 406 691 370 760 858 042 752 96 × 2 = 0 + 0.000 002 813 382 741 521 716 085 505 92;
  • 8) 0.000 002 813 382 741 521 716 085 505 92 × 2 = 0 + 0.000 005 626 765 483 043 432 171 011 84;
  • 9) 0.000 005 626 765 483 043 432 171 011 84 × 2 = 0 + 0.000 011 253 530 966 086 864 342 023 68;
  • 10) 0.000 011 253 530 966 086 864 342 023 68 × 2 = 0 + 0.000 022 507 061 932 173 728 684 047 36;
  • 11) 0.000 022 507 061 932 173 728 684 047 36 × 2 = 0 + 0.000 045 014 123 864 347 457 368 094 72;
  • 12) 0.000 045 014 123 864 347 457 368 094 72 × 2 = 0 + 0.000 090 028 247 728 694 914 736 189 44;
  • 13) 0.000 090 028 247 728 694 914 736 189 44 × 2 = 0 + 0.000 180 056 495 457 389 829 472 378 88;
  • 14) 0.000 180 056 495 457 389 829 472 378 88 × 2 = 0 + 0.000 360 112 990 914 779 658 944 757 76;
  • 15) 0.000 360 112 990 914 779 658 944 757 76 × 2 = 0 + 0.000 720 225 981 829 559 317 889 515 52;
  • 16) 0.000 720 225 981 829 559 317 889 515 52 × 2 = 0 + 0.001 440 451 963 659 118 635 779 031 04;
  • 17) 0.001 440 451 963 659 118 635 779 031 04 × 2 = 0 + 0.002 880 903 927 318 237 271 558 062 08;
  • 18) 0.002 880 903 927 318 237 271 558 062 08 × 2 = 0 + 0.005 761 807 854 636 474 543 116 124 16;
  • 19) 0.005 761 807 854 636 474 543 116 124 16 × 2 = 0 + 0.011 523 615 709 272 949 086 232 248 32;
  • 20) 0.011 523 615 709 272 949 086 232 248 32 × 2 = 0 + 0.023 047 231 418 545 898 172 464 496 64;
  • 21) 0.023 047 231 418 545 898 172 464 496 64 × 2 = 0 + 0.046 094 462 837 091 796 344 928 993 28;
  • 22) 0.046 094 462 837 091 796 344 928 993 28 × 2 = 0 + 0.092 188 925 674 183 592 689 857 986 56;
  • 23) 0.092 188 925 674 183 592 689 857 986 56 × 2 = 0 + 0.184 377 851 348 367 185 379 715 973 12;
  • 24) 0.184 377 851 348 367 185 379 715 973 12 × 2 = 0 + 0.368 755 702 696 734 370 759 431 946 24;
  • 25) 0.368 755 702 696 734 370 759 431 946 24 × 2 = 0 + 0.737 511 405 393 468 741 518 863 892 48;
  • 26) 0.737 511 405 393 468 741 518 863 892 48 × 2 = 1 + 0.475 022 810 786 937 483 037 727 784 96;
  • 27) 0.475 022 810 786 937 483 037 727 784 96 × 2 = 0 + 0.950 045 621 573 874 966 075 455 569 92;
  • 28) 0.950 045 621 573 874 966 075 455 569 92 × 2 = 1 + 0.900 091 243 147 749 932 150 911 139 84;
  • 29) 0.900 091 243 147 749 932 150 911 139 84 × 2 = 1 + 0.800 182 486 295 499 864 301 822 279 68;
  • 30) 0.800 182 486 295 499 864 301 822 279 68 × 2 = 1 + 0.600 364 972 590 999 728 603 644 559 36;
  • 31) 0.600 364 972 590 999 728 603 644 559 36 × 2 = 1 + 0.200 729 945 181 999 457 207 289 118 72;
  • 32) 0.200 729 945 181 999 457 207 289 118 72 × 2 = 0 + 0.401 459 890 363 998 914 414 578 237 44;
  • 33) 0.401 459 890 363 998 914 414 578 237 44 × 2 = 0 + 0.802 919 780 727 997 828 829 156 474 88;
  • 34) 0.802 919 780 727 997 828 829 156 474 88 × 2 = 1 + 0.605 839 561 455 995 657 658 312 949 76;
  • 35) 0.605 839 561 455 995 657 658 312 949 76 × 2 = 1 + 0.211 679 122 911 991 315 316 625 899 52;
  • 36) 0.211 679 122 911 991 315 316 625 899 52 × 2 = 0 + 0.423 358 245 823 982 630 633 251 799 04;
  • 37) 0.423 358 245 823 982 630 633 251 799 04 × 2 = 0 + 0.846 716 491 647 965 261 266 503 598 08;
  • 38) 0.846 716 491 647 965 261 266 503 598 08 × 2 = 1 + 0.693 432 983 295 930 522 533 007 196 16;
  • 39) 0.693 432 983 295 930 522 533 007 196 16 × 2 = 1 + 0.386 865 966 591 861 045 066 014 392 32;
  • 40) 0.386 865 966 591 861 045 066 014 392 32 × 2 = 0 + 0.773 731 933 183 722 090 132 028 784 64;
  • 41) 0.773 731 933 183 722 090 132 028 784 64 × 2 = 1 + 0.547 463 866 367 444 180 264 057 569 28;
  • 42) 0.547 463 866 367 444 180 264 057 569 28 × 2 = 1 + 0.094 927 732 734 888 360 528 115 138 56;
  • 43) 0.094 927 732 734 888 360 528 115 138 56 × 2 = 0 + 0.189 855 465 469 776 721 056 230 277 12;
  • 44) 0.189 855 465 469 776 721 056 230 277 12 × 2 = 0 + 0.379 710 930 939 553 442 112 460 554 24;
  • 45) 0.379 710 930 939 553 442 112 460 554 24 × 2 = 0 + 0.759 421 861 879 106 884 224 921 108 48;
  • 46) 0.759 421 861 879 106 884 224 921 108 48 × 2 = 1 + 0.518 843 723 758 213 768 449 842 216 96;
  • 47) 0.518 843 723 758 213 768 449 842 216 96 × 2 = 1 + 0.037 687 447 516 427 536 899 684 433 92;
  • 48) 0.037 687 447 516 427 536 899 684 433 92 × 2 = 0 + 0.075 374 895 032 855 073 799 368 867 84;
  • 49) 0.075 374 895 032 855 073 799 368 867 84 × 2 = 0 + 0.150 749 790 065 710 147 598 737 735 68;
  • 50) 0.150 749 790 065 710 147 598 737 735 68 × 2 = 0 + 0.301 499 580 131 420 295 197 475 471 36;
  • 51) 0.301 499 580 131 420 295 197 475 471 36 × 2 = 0 + 0.602 999 160 262 840 590 394 950 942 72;
  • 52) 0.602 999 160 262 840 590 394 950 942 72 × 2 = 1 + 0.205 998 320 525 681 180 789 901 885 44;
  • 53) 0.205 998 320 525 681 180 789 901 885 44 × 2 = 0 + 0.411 996 641 051 362 361 579 803 770 88;
  • 54) 0.411 996 641 051 362 361 579 803 770 88 × 2 = 0 + 0.823 993 282 102 724 723 159 607 541 76;
  • 55) 0.823 993 282 102 724 723 159 607 541 76 × 2 = 1 + 0.647 986 564 205 449 446 319 215 083 52;
  • 56) 0.647 986 564 205 449 446 319 215 083 52 × 2 = 1 + 0.295 973 128 410 898 892 638 430 167 04;
  • 57) 0.295 973 128 410 898 892 638 430 167 04 × 2 = 0 + 0.591 946 256 821 797 785 276 860 334 08;
  • 58) 0.591 946 256 821 797 785 276 860 334 08 × 2 = 1 + 0.183 892 513 643 595 570 553 720 668 16;
  • 59) 0.183 892 513 643 595 570 553 720 668 16 × 2 = 0 + 0.367 785 027 287 191 141 107 441 336 32;
  • 60) 0.367 785 027 287 191 141 107 441 336 32 × 2 = 0 + 0.735 570 054 574 382 282 214 882 672 64;
  • 61) 0.735 570 054 574 382 282 214 882 672 64 × 2 = 1 + 0.471 140 109 148 764 564 429 765 345 28;
  • 62) 0.471 140 109 148 764 564 429 765 345 28 × 2 = 0 + 0.942 280 218 297 529 128 859 530 690 56;
  • 63) 0.942 280 218 297 529 128 859 530 690 56 × 2 = 1 + 0.884 560 436 595 058 257 719 061 381 12;
  • 64) 0.884 560 436 595 058 257 719 061 381 12 × 2 = 1 + 0.769 120 873 190 116 515 438 122 762 24;
  • 65) 0.769 120 873 190 116 515 438 122 762 24 × 2 = 1 + 0.538 241 746 380 233 030 876 245 524 48;
  • 66) 0.538 241 746 380 233 030 876 245 524 48 × 2 = 1 + 0.076 483 492 760 466 061 752 491 048 96;
  • 67) 0.076 483 492 760 466 061 752 491 048 96 × 2 = 0 + 0.152 966 985 520 932 123 504 982 097 92;
  • 68) 0.152 966 985 520 932 123 504 982 097 92 × 2 = 0 + 0.305 933 971 041 864 247 009 964 195 84;
  • 69) 0.305 933 971 041 864 247 009 964 195 84 × 2 = 0 + 0.611 867 942 083 728 494 019 928 391 68;
  • 70) 0.611 867 942 083 728 494 019 928 391 68 × 2 = 1 + 0.223 735 884 167 456 988 039 856 783 36;
  • 71) 0.223 735 884 167 456 988 039 856 783 36 × 2 = 0 + 0.447 471 768 334 913 976 079 713 566 72;
  • 72) 0.447 471 768 334 913 976 079 713 566 72 × 2 = 0 + 0.894 943 536 669 827 952 159 427 133 44;
  • 73) 0.894 943 536 669 827 952 159 427 133 44 × 2 = 1 + 0.789 887 073 339 655 904 318 854 266 88;
  • 74) 0.789 887 073 339 655 904 318 854 266 88 × 2 = 1 + 0.579 774 146 679 311 808 637 708 533 76;
  • 75) 0.579 774 146 679 311 808 637 708 533 76 × 2 = 1 + 0.159 548 293 358 623 617 275 417 067 52;
  • 76) 0.159 548 293 358 623 617 275 417 067 52 × 2 = 0 + 0.319 096 586 717 247 234 550 834 135 04;
  • 77) 0.319 096 586 717 247 234 550 834 135 04 × 2 = 0 + 0.638 193 173 434 494 469 101 668 270 08;
  • 78) 0.638 193 173 434 494 469 101 668 270 08 × 2 = 1 + 0.276 386 346 868 988 938 203 336 540 16;

We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit) and at least one integer that was different from zero => FULL STOP (Losing precision - the converted number we get in the end will be just a very good approximation of the initial one).


4. Construct the base 2 representation of the fractional part of the number.

Take all the integer parts of the multiplying operations, starting from the top of the constructed list above:


0.000 000 021 979 552 668 138 406 918 015(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0011 0100 1011 1100 0100 1110 01(2)

5. Positive number before normalization:

0.000 000 021 979 552 668 138 406 918 015(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0011 0100 1011 1100 0100 1110 01(2)

6. Normalize the binary representation of the number.

Shift the decimal mark 26 positions to the right, so that only one non zero digit remains to the left of it:


0.000 000 021 979 552 668 138 406 918 015(10) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0011 0100 1011 1100 0100 1110 01(2) =

0.0000 0000 0000 0000 0000 0000 0101 1110 0110 0110 1100 0110 0001 0011 0100 1011 1100 0100 1110 01(2) × 20 =

1.0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001(2) × 2-26


7. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

Sign 0 (a positive number)

Exponent (unadjusted): -26

Mantissa (not normalized):
1.0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001


8. Adjust the exponent.

Use the 11 bit excess/bias notation:


Exponent (adjusted) =

Exponent (unadjusted) + 2(11-1) - 1 =

-26 + 2(11-1) - 1 =

(-26 + 1 023)(10) =

997(10)


9. Convert the adjusted exponent from the decimal (base 10) to 11 bit binary.

Use the same technique of repeatedly dividing by 2:


  • division = quotient + remainder;
  • 997 ÷ 2 = 498 + 1;
  • 498 ÷ 2 = 249 + 0;
  • 249 ÷ 2 = 124 + 1;
  • 124 ÷ 2 = 62 + 0;
  • 62 ÷ 2 = 31 + 0;
  • 31 ÷ 2 = 15 + 1;
  • 15 ÷ 2 = 7 + 1;
  • 7 ÷ 2 = 3 + 1;
  • 3 ÷ 2 = 1 + 1;
  • 1 ÷ 2 = 0 + 1;

10. Construct the base 2 representation of the adjusted exponent.

Take all the remainders starting from the bottom of the list constructed above.


Exponent (adjusted) =

997(10) =

011 1110 0101(2)


11. Normalize the mantissa.

a) Remove the leading (the leftmost) bit, since it's allways 1, and the decimal point, if the case.

b) Adjust its length to 52 bits, only if necessary (not the case here).


Mantissa (normalized) =

1. 0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001 =

0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001


12. The three elements that make up the number's 64 bit double precision IEEE 754 binary floating point representation:

Sign (1 bit) =
0 (a positive number)

Exponent (11 bits) =
011 1110 0101

Mantissa (52 bits) =
0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001


Decimal number 0.000 000 021 979 552 668 138 406 918 015 converted to 64 bit double precision IEEE 754 binary floating point representation:

0 - 011 1110 0101 - 0111 1001 1001 1011 0001 1000 0100 1101 0010 1111 0001 0011 1001

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How to convert numbers from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point standard

Follow the steps below to convert a base 10 decimal number to 64 bit double precision IEEE 754 binary floating point:

  • 1. If the number to be converted is negative, start with its the positive version.
  • 2. First convert the integer part. Divide repeatedly by 2 the positive representation of the integer number that is to be converted to binary, until we get a quotient that is equal to zero, keeping track of each remainder.
  • 3. Construct the base 2 representation of the positive integer part of the number, by taking all the remainders from the previous operations, starting from the bottom of the list constructed above. Thus, the last remainder of the divisions becomes the first symbol (the leftmost) of the base two number, while the first remainder becomes the last symbol (the rightmost).
  • 4. Then convert the fractional part. Multiply the number repeatedly by 2, until we get a fractional part that is equal to zero, keeping track of each integer part of the results.
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the multiplying operations, starting from the top of the list constructed above (they should appear in the binary representation, from left to right, in the order they have been calculated).
  • 6. Normalize the binary representation of the number, shifting the decimal mark (the decimal point) "n" positions either to the left, or to the right, so that only one non zero digit remains to the left of the decimal mark.
  • 7. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary, by using the same technique of repeatedly dividing by 2, as shown above:
    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1
  • 8. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal mark, if the case) and adjust its length to 52 bits, either by removing the excess bits from the right (losing precision...) or by adding extra bits set on '0' to the right.
  • 9. Sign (it takes 1 bit) is either 1 for a negative or 0 for a positive number.

Example: convert the negative number -31.640 215 from the decimal system (base ten) to 64 bit double precision IEEE 754 binary floating point:

  • 1. Start with the positive version of the number:

    |-31.640 215| = 31.640 215

  • 2. First convert the integer part, 31. Divide it repeatedly by 2, keeping track of each remainder, until we get a quotient that is equal to zero:
    • division = quotient + remainder;
    • 31 ÷ 2 = 15 + 1;
    • 15 ÷ 2 = 7 + 1;
    • 7 ÷ 2 = 3 + 1;
    • 3 ÷ 2 = 1 + 1;
    • 1 ÷ 2 = 0 + 1;
    • We have encountered a quotient that is ZERO => FULL STOP
  • 3. Construct the base 2 representation of the integer part of the number by taking all the remainders of the previous dividing operations, starting from the bottom of the list constructed above:

    31(10) = 1 1111(2)

  • 4. Then, convert the fractional part, 0.640 215. Multiply repeatedly by 2, keeping track of each integer part of the results, until we get a fractional part that is equal to zero:
    • #) multiplying = integer + fractional part;
    • 1) 0.640 215 × 2 = 1 + 0.280 43;
    • 2) 0.280 43 × 2 = 0 + 0.560 86;
    • 3) 0.560 86 × 2 = 1 + 0.121 72;
    • 4) 0.121 72 × 2 = 0 + 0.243 44;
    • 5) 0.243 44 × 2 = 0 + 0.486 88;
    • 6) 0.486 88 × 2 = 0 + 0.973 76;
    • 7) 0.973 76 × 2 = 1 + 0.947 52;
    • 8) 0.947 52 × 2 = 1 + 0.895 04;
    • 9) 0.895 04 × 2 = 1 + 0.790 08;
    • 10) 0.790 08 × 2 = 1 + 0.580 16;
    • 11) 0.580 16 × 2 = 1 + 0.160 32;
    • 12) 0.160 32 × 2 = 0 + 0.320 64;
    • 13) 0.320 64 × 2 = 0 + 0.641 28;
    • 14) 0.641 28 × 2 = 1 + 0.282 56;
    • 15) 0.282 56 × 2 = 0 + 0.565 12;
    • 16) 0.565 12 × 2 = 1 + 0.130 24;
    • 17) 0.130 24 × 2 = 0 + 0.260 48;
    • 18) 0.260 48 × 2 = 0 + 0.520 96;
    • 19) 0.520 96 × 2 = 1 + 0.041 92;
    • 20) 0.041 92 × 2 = 0 + 0.083 84;
    • 21) 0.083 84 × 2 = 0 + 0.167 68;
    • 22) 0.167 68 × 2 = 0 + 0.335 36;
    • 23) 0.335 36 × 2 = 0 + 0.670 72;
    • 24) 0.670 72 × 2 = 1 + 0.341 44;
    • 25) 0.341 44 × 2 = 0 + 0.682 88;
    • 26) 0.682 88 × 2 = 1 + 0.365 76;
    • 27) 0.365 76 × 2 = 0 + 0.731 52;
    • 28) 0.731 52 × 2 = 1 + 0.463 04;
    • 29) 0.463 04 × 2 = 0 + 0.926 08;
    • 30) 0.926 08 × 2 = 1 + 0.852 16;
    • 31) 0.852 16 × 2 = 1 + 0.704 32;
    • 32) 0.704 32 × 2 = 1 + 0.408 64;
    • 33) 0.408 64 × 2 = 0 + 0.817 28;
    • 34) 0.817 28 × 2 = 1 + 0.634 56;
    • 35) 0.634 56 × 2 = 1 + 0.269 12;
    • 36) 0.269 12 × 2 = 0 + 0.538 24;
    • 37) 0.538 24 × 2 = 1 + 0.076 48;
    • 38) 0.076 48 × 2 = 0 + 0.152 96;
    • 39) 0.152 96 × 2 = 0 + 0.305 92;
    • 40) 0.305 92 × 2 = 0 + 0.611 84;
    • 41) 0.611 84 × 2 = 1 + 0.223 68;
    • 42) 0.223 68 × 2 = 0 + 0.447 36;
    • 43) 0.447 36 × 2 = 0 + 0.894 72;
    • 44) 0.894 72 × 2 = 1 + 0.789 44;
    • 45) 0.789 44 × 2 = 1 + 0.578 88;
    • 46) 0.578 88 × 2 = 1 + 0.157 76;
    • 47) 0.157 76 × 2 = 0 + 0.315 52;
    • 48) 0.315 52 × 2 = 0 + 0.631 04;
    • 49) 0.631 04 × 2 = 1 + 0.262 08;
    • 50) 0.262 08 × 2 = 0 + 0.524 16;
    • 51) 0.524 16 × 2 = 1 + 0.048 32;
    • 52) 0.048 32 × 2 = 0 + 0.096 64;
    • 53) 0.096 64 × 2 = 0 + 0.193 28;
    • We didn't get any fractional part that was equal to zero. But we had enough iterations (over Mantissa limit = 52) and at least one integer part that was different from zero => FULL STOP (losing precision...).
  • 5. Construct the base 2 representation of the fractional part of the number, by taking all the integer parts of the previous multiplying operations, starting from the top of the constructed list above:

    0.640 215(10) = 0.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 6. Summarizing - the positive number before normalization:

    31.640 215(10) = 1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2)

  • 7. Normalize the binary representation of the number, shifting the decimal mark 4 positions to the left so that only one non-zero digit stays to the left of the decimal mark:

    31.640 215(10) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) =
    1 1111.1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 20 =
    1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0(2) × 24

  • 8. Up to this moment, there are the following elements that would feed into the 64 bit double precision IEEE 754 binary floating point representation:

    Sign: 1 (a negative number)

    Exponent (unadjusted): 4

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

  • 9. Adjust the exponent in 11 bit excess/bias notation and then convert it from decimal (base 10) to 11 bit binary (base 2), by using the same technique of repeatedly dividing it by 2, as shown above:

    Exponent (adjusted) = Exponent (unadjusted) + 2(11-1) - 1 = (4 + 1023)(10) = 1027(10) =
    100 0000 0011(2)

  • 10. Normalize mantissa, remove the leading (leftmost) bit, since it's allways '1' (and the decimal sign) and adjust its length to 52 bits, by removing the excess bits, from the right (losing precision...):

    Mantissa (not-normalized): 1.1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100 1010 0

    Mantissa (normalized): 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Conclusion:

    Sign (1 bit) = 1 (a negative number)

    Exponent (8 bits) = 100 0000 0011

    Mantissa (52 bits) = 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100

  • Number -31.640 215, converted from decimal system (base 10) to 64 bit double precision IEEE 754 binary floating point =
    1 - 100 0000 0011 - 1111 1010 0011 1110 0101 0010 0001 0101 0111 0110 1000 1001 1100